Most proofs that i see, use Taylor expansion upto second order and ignore higher order terms and make use of first order necessary condition. Then the claim is if Hessian is Positive definite then then value of function is only going to increase in the neighborhood of point under consideration hence it has to be strict local minima.
But then isn't this a possibility that further term that we have ignored from taylor expansion may be sum to negative of the second order term and then we will have same value of the function in the neighborhood of the point which we otherwise conclude as strict local minima.
What is wrong in my thinking or what am i missing here ?
What you need to take into account is Taylor's Theorem itself, i.e. you can write any smooth function as a Taylor sum plus remainder term. You can analyse this remainder term and find an explicit version of it, see e.g.
https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder
If you now assume $f''(x_0)>0$, then by continuity you also have $f''(x)>0$ in a neighbourhood of $x_0$. Hence the remainder term has a sign and therefore you get that $x_0$ with $f'(x_0)=0$ is a local extremum.